

A152466


a(1) = 252, a(n) is a(n1) multiplied by the smallest prime factor of a(n1)+1.


4



252, 2772, 130284, 651420, 219528540, 257067920340, 4370154645780, 292800361267260, 11023640801351071740, 13475008472558425746927448860, 5107028211099643358085503117940, 1313771981231475489737485570488833367540
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OFFSET

1,1


COMMENTS

Conjecture: this sequence contains no terms k where k+1 is prime. (All similar sequences that start with numbers less than 252 are known to contain terms k where k+1 is prime.)
The next few similar sequences that seem to have this property are those that begin with a(1) = 322, 622, 664, 776, and 830.  J. Lowell, Mar 25 2014
Adding 1 to the 30th term of this sequence gives a 152digit composite number with no factors found in ECM after hundreds of curves.  J. Lowell, Jan 11 2022
The above conjecture has now been disproved: adding 1 to the 39th term of this sequence gives a prime number.  Andrea Concaro, Dec 31 2022
Calculating a(114) requires partial factorization of a(113)+1, a 1022digit composite number.  Tyler Busby, Jan 21 2023


LINKS

Tyler Busby, Conjectured Table of n, a(n) for n = 1..113 using ellipticcurve factorization to obtain probable smallest factors when other methods were computationally unfeasible.


FORMULA



EXAMPLE

First term is 252. Smallest prime factor of 253 is 11, so next term is 252 * 11 = 2772.


MATHEMATICA

a = {252}; Do[AppendTo[a, a[[ 1]]*FactorInteger[a[[ 1]] + 1][[1, 1]]], {10}]; a (* Stefan Steinerberger, Dec 06 2008 *)
NestList[#*FactorInteger[#+1][[1, 1]]&, 252, 20] (* Harvey P. Dale, Apr 03 2015 *)


PROG

(PARI) findsmallestfactor(n)=if(isprime(n), n, forprime(p=2, 1e6, if(n%p==0, return(p))); factor(n)[1, 1])
lista(n)={vals=Vec([252], n); for(i=2, n, vals[i]=findsmallestfactor(vals[i1]+1)*vals[i1]); vals} \\ Tyler Busby, Jan 14 2023


CROSSREFS

Cf. A020639 (smallest prime factor), A238584 (ratios of consecutive terms).


KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



